Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that

$$\frac{1}{2}-\frac{1}{2e}<\int_0^{+\infty}e^{-x^2}dx<1+\frac{1}{2e}$$

I know that one way to do this is to evaluate the integral in the middle, and then compare these three numbers. I wonder how can we do this without explicitly compute the integral?

share|improve this question
    
Where can the e in the bound possible come from i wonder... –  Lost1 Jan 4 at 1:20
    
Integration by parts would seem to be one of the many magic bullets... –  Igor Rivin Jan 4 at 1:27

1 Answer 1

up vote 7 down vote accepted

Proof for the upper bound: $$\int_0^{\infty}e^{-x^2}dx =\int_{0}^{1}e^{-x^2}dx+\int_{1}^{\infty} e^{-x^2} dx <\int_{0}^{1}1 dx+\int_{1}^{\infty} x e^{-x^2} dx =1+\frac{1}{2e}$$ where the last integral is easily calculated by substituting $u=-x^2$.


Proof for the lower bound: $$\int_0^{\infty}e^{-x^2}dx > \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1} xe^{-x^2} dx =\frac{1}{2}-\frac{1}{2e}$$ where the last integral is easily calculated by substituting $u=-x^2$.

share|improve this answer
    
Oh nice. This is actually quite straight forward... –  Lost1 Jan 4 at 1:30
    
This is pretty awesome! Thanks!Really like this proof –  Steven Jan 4 at 1:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.